Evaluation of Limits MCQ Quiz - Objective Question with Answer for Evaluation of Limits - Download Free PDF

Calculation:

Given,

The function is \( f(x) = x^2 + 9 \).

Statement I: f(x) is an increasing function.

The derivative of f(x) is:

\( f'(x) = \frac{d}{dx}(x^2 + 9) = 2x \)

When \(( x > 0 )\), \(( f'(x) > 0 )\), so (f(x) is increasing.

When (x < 0), f'(x) < 0 ), so f(x)  is decreasing.

At (x = 0), ( f'(x) = 0 ), meaning the function is neither increasing nor decreasing at this point.

Hence, f(x)  is not entirely increasing. It is increasing for (x > 0) and decreasing for ( x < 0).

Statement II: f(x) has local maximum at x = 0

Since the function\( f(x) = x^2 + 9 \) is a parabola opening upwards (because the coefficient of x² is positive), it has a global minimum at x = 0, not a local maximum.

Conclusion:

- Statement I is incorrect because the function is not entirely increasing. It is increasing for x > 0  and decreasing for  x < 0 .

- Statement II is incorrect because the function has a global minimum at x = 0, not a local maximum.

Hence, the correct answer is Option 4. 

Calculation:

Given,

The function is \( f(x) = \sqrt{x^2 + 9} - 3 \) and \( g(x) = \sqrt{x^2 + 16} - 4 \).

We are tasked with finding:

\( \lim_{x \to 0} \frac{f(x)}{g(x)} \)

Multiply both the numerator and denominator by their respective conjugates:

\( \frac{\sqrt{x^2 + 9} - 3}{\sqrt{x^2 + 16} - 4} \times \frac{\sqrt{x^2 + 9} + 3}{\sqrt{x^2 + 9} + 3} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 16} + 4} \)

Simplify the numerator:

\( (\sqrt{x^2 + 9} - 3)(\sqrt{x^2 + 9} + 3) = x^2 \)

Simplify the denominator:

\( (\sqrt{x^2 + 16} - 4)(\sqrt{x^2 + 16} + 4) = x^2 \)

Now, the expression becomes:

\( \frac{x^2}{x^2} \times \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \)

Simplify and evaluate the limit:

\( \frac{\sqrt{x^2 + 16} + 4}{\sqrt{x^2 + 9} + 3} \) becomes:

\( \frac{\sqrt{16} + 4}{\sqrt{9} + 3} = \frac{4 + 4}{3 + 3} = \frac{8}{6} = \frac{4}{3} \)

Hence, the correct answer is Option 3.

Calculation:

\(\rm \displaystyle \Rightarrow \lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}=\lim _{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\sin 4 x \cdot \cos 6 x}{\sin 6 x \cdot \cos 4 x}}=e^{2 / 3}\)

\(\rm \displaystyle \Rightarrow \lim _{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^{\left.\frac{\lambda}{\cos x} \right\rvert\,}=e^{\lambda}\)

⇒ f(π/2) = µ 

For continuous function ⇒ e^2/3 = e^λ = µ 

\(\lambda=\frac{2}{3}, \mu=\mathrm{e}^{2 / 3}\)

Now, 6λ + 6log_eµ + µ⁶ – e^6λ = 10

Hence, the correct answer is Option 4. 

Answer (1)

Sol.

\(\lim _{x \rightarrow 0} \frac{\cos 2 x+a \cos ^{4} x-b}{x^{4}}=L\)

\(\lim _{x \rightarrow 0} \frac{2 \cos ^{2} x-1+a \cos ^{4} x-b}{x^{4}}=L \quad \ldots(1)\)

To get the finite value,

1 + a – b = 0 

⇒ \(a=b-1 \quad \quad ...(2)\)

Apply L Hospital 

\(\lim _{x \rightarrow 0} \frac{4 \cos x(-\sin x)+4 a \cos ^{3} x(-\sin x)}{4 x^{3}}\)

\(\lim _{x \rightarrow 0} \frac{4 \cos x+4 a \cos ^{3} x}{4 x^{3}}\left(\frac{-\sin x}{x}\right)\)

To get the finite value, a = –1

Also from (1)

b = 0 

∴ a = b = -1

Calcu;ation: 

 \(\rm \operatorname{Lim}_{n \rightarrow \infty} \frac{0+3+6+9+\ldots \ldots n \text { terms }}{\sqrt{2 n^{4}+4 n+3}-\sqrt{n^{4}+5 n+4}}\)

⇒ \(\rm \operatorname{Lim}_{n \rightarrow \infty}\rm \frac{3 n(n-1)}{2\left(\sqrt{2 n^{4}+4 n+3}-\sqrt{n^{4}+5 n+4}\right)}\)

= \(\frac{3}{2(\sqrt{2}-1)}=\frac{3}{2}(\sqrt{2}+1)\)

Hence, the correct answer is Option 3.

Concept:

• 1 - cos 2θ = 2 sin2 θ
• \(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{\sin x}}{x} = 1\)

Calculation:

\(\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {1 - \cos2x} \right)}^2}\;}}{{{x^4}}}\)

= \(\;\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{{{\left( {2{{\sin }^2}x} \right)}^2}}}{{{x^4}}}\)          (1 - cos 2θ = 2 sin² θ)

= \(\;\mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{4{{\sin }^4}x}}{{{x^4}}}\)

= \(\mathop {\lim }\limits_{x \to 0} 4\; × \;{\left( {\frac{{\sin x}}{x}} \right)^4}\)

= 4 × 1 = 4

Concept:

\(\rm \mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+x)}}{x}} = 1\)

Calculation:

\(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x}\)

\(\rm = \mathop {\lim }\limits_{x\rightarrow 0} \frac{\frac{\log (1+2x)}{2x} \times 2x}{\frac{\tan 2x}{2x} \times 2x}\\= \frac{\mathop {\lim }\limits_{x\rightarrow 0}\frac{\log (1+2x)}{2x} }{\mathop {\lim }\limits_{x\rightarrow 0}\frac{\tan 2x}{2x} }\)

As we know \(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\) and \(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+x)}}{x}} = 1\)

Therefore, \(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan 2x}}{2x}} = 1\) and \(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\log (1+2x)}}{2x}} = 1\)

Hence \(\rm \mathop {\lim }\limits_{x\rightarrow 0} \frac{\log (1+2x)}{\tan 2x} = \frac 1 1=1\)

Calculation:

We have to find the value of \(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)

\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)       [Form \(\frac{∞}{∞}\)]

This limit is of the form \(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞  out of the numerator and denominator.

\(\rm \mathop {\lim }\limits_{x\rightarrow ∞} \frac{x}{\sqrt{1+2x^2}}\)

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x}{x\sqrt{\frac{1}{x^2}+2}}\)

Factor x becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\sqrt{\frac{1}{x^2}+2}}\)

= \(\frac{1}{\sqrt{\frac{1}{\infty^2}+2}}=\frac{1}{\sqrt{0+2}}=\frac{1}{\sqrt 2}\)

Calculation:

We have to find the value of \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)

\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)       [Form \(\frac{∞}{∞}\)]

This limit is of the form \(\frac{∞}{∞}\), Here, We can cancel a factor going to ∞  out of the numerator and denominator.

\(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{{1+x^2}}\)

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{x^2}{x^2\left({\frac {1}{x^2}+1}\right)}\)

Factor x² becomes ∞ at x tends to ∞, So we need to cancel this factor from numerator and denominator.

= \(\rm \mathop {\lim }\limits_{x\rightarrow \infty} \frac{1}{\left({\frac {1}{x^2}+1}\right)}\)

= \(\frac{1}{{\frac{1}{\infty^2}+1}}=\frac{1}{{0+1}}=1\)

Concept:​

• \(\rm \displaystyle \lim_{x\to0}\dfrac{\tan x}{x}=1\).
• \(\rm \dfrac{d}{dx}\tan x=\sec^2x\).
• \(\rm \dfrac{d}{dx}\sec x=\tan x\sec x\).
• \(\rm \dfrac{d}{dx}\left[f(x)\times g(x)\right]=f(x)\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)\).

Indeterminate Forms: Any expression whose value cannot be defined, like \(\dfrac00\), \(\pm\dfrac{\infty}{\infty}\), 0⁰, ∞⁰ etc.

• For the indeterminate form \(\dfrac 0 0\), first try to rationalize by multiplying with the conjugate, or simplify by cancelling some terms in the numerator and denominator. Else, use the L'Hospital's rule.
• L'Hospital's Rule: For the differentiable functions f(x) and g(x), the \(\rm \displaystyle \lim_{x\to c} \dfrac{f(x)}{g(x)}\), if f(x) and g(x) are both 0 or ±∞ (i.e. an Indeterminate Form) is equal to the \(\rm \displaystyle \lim_{x\to c} \dfrac{f'(x)}{g'(x)}\) if it exists.

Calculation:

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\dfrac00\) is an indeterminate form. Let us simplify and use the L'Hospital's Rule.

\(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=\lim_{x\to 0}\left[\dfrac{\tan x - x}{x^3}\times\dfrac{x}{\tan x}\right]\).

We know that \(\rm \displaystyle\lim_{x\to 0}\dfrac{x}{\tan x}=1\), but \(\rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}\) is still an indeterminate form, so we use L'Hospital's Rule:

\(\rm \displaystyle \rm \displaystyle \lim_{x\to 0}\dfrac{\tan x - x}{x^3}=\lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x - 1}{3x^2}= \lim_{x\to 0}\dfrac{2\sec x(\sec x\tan x)}{6x}=\lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}\), which is still an indeterminate form, so we use L'Hospital's Rule again:

\(\rm \displaystyle \lim_{x\to 0}\dfrac{\sec^2 x\tan x}{3x}=\lim_{x\to 0}\dfrac{\sec^2 x\sec^2 x+\tan x[2\sec x(\sec x \tan x)]}{3}=\dfrac{1}{3}\).

∴ \(\rm \displaystyle\lim_{x\to 0}\dfrac{\tan x - x}{x^2 \tan x}=1\times\dfrac{1}{3}=\dfrac{1}{3}\).

Concept:

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac { (a^x - 1) }{x} = \log a\)

log mⁿ = n log m

Calculation:

\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0\)

Formula used:

\(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{sin\ x}{x}}\right)=1\)

Calculation:

\(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{\sqrt{1-cosx^2}}{(1-cosx)}}\right)\)

Since, 1 - cos 2θ = sin²θ

⇒ \(\rm \displaystyle\lim_ {x\rightarrow0}\left({\frac{√{2sin^2\frac{x^2}{2}}}{(2sin^2\frac{x}{2})}}\right)\)

⇒ \(\frac{1}{√ 2}\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{{\frac{x^2}{2}\times sin\frac{x^2}{2}}}{(sin^2\frac{x}{2})\times\frac{x^2}{2}}}\right)\)

∴  \(\frac{2}{√ 2}\rm \displaystyle\lim_ {x\rightarrow 0}\left(\frac{sin\frac{x^2}{2}}{\frac{x^2}{2}}\right)\times \left(\frac{\frac{x}{2}}{sin\frac{x}{2}}\right)^2\) = √2

Concept:

\(\rm \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\)

Calculation:

\(\rm \displaystyle \lim_{x → ∞} x \sin \left(\frac{π} {x}\right)\)

= \(\rm \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{π} {x}\right)}{\left(\frac{1}{x} \right )}\)

= \(\rm \displaystyle \lim_{x → ∞} \frac{\sin \left(\frac{π} {x}\right)}{\left(\frac{π}{x} \right )} × π\)

Let \(\rm \frac {π}{x} = t\)

If x → ∞ then t → 0

= \(\rm \displaystyle \lim_{t \rightarrow 0} \frac{\sin t}{t} × π\)

= 1 × π 

= π 

Concept:

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} = 1\\\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+x) }{x}\)

Calculation:

We have to find the value of \(\rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}\)

As we know, \(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) \cdot g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) \cdot \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\therefore \rm \displaystyle\lim_{x\rightarrow 0} \dfrac{\sin x \log (1-x)}{x^2}= \rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\sin x }{x} × ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1-x) }{x}\)

= 1 × \(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{x}\)

= \(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{-(-x)}\)

= \(-1 × \rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac {\log (1+(-x)) }{(-x)}\)

= -1 × 1

= -1

\(\mathop {\lim }\limits_{n \to \infty } \frac{{{2^{n + 1}} + {3^{n + 1}}}}{{{2^n} + {3^n}}} \)

This can be written as:

\(= \mathop {\lim }\limits_{n \to \infty } \frac{{{2^n}2 + {3^n}3}}{{{2^n} + {3^n}}}\)

Taking 3ⁿ common, we can write:

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{{3^n}\left[ {2.{{\left( {\frac{2}{3}} \right)}^n} + 3} \right]}}{{{3^n}\left[ {{{\left( {\frac{2}{3}} \right)}^n} + 1} \right]}}\)

\( = \mathop {\lim }\limits_{n \to \infty } \frac{{2.{{\left( {\frac{2}{3}} \right)}^n} + 3}}{{\left[ {{{\left( {\frac{2}{3}} \right)}^n} + 1} \right]}}\)

Here \(\frac 2 3 < 1\)

So, \(\left[ \frac {2}{3}\right]^{\infty} = 0\)

\(= \frac{{0 + 3}}{{0 + 1}} = 3\)